NEET UG 2021 — Chemistry Physical Chemistry

2021mcqmedium

The following solutions were prepared by dissolving $10g$ of glucose $({C}_{6}{H}_{12}{O}_{6})$ in $250\mathrm{mL}$ of water $({P}_{1}),10g$ of urea $({\mathrm{CH}}_{4}{N}_{2}O)$ in $250\mathrm{mL}$ of water $({P}_{2})$ and $10g$ of sucrose $({C}_{12}{H}_{22}{O}_{11})$ in $250\mathrm{mL}$ of water $({P}_{3}).$ The right option for the decreasing order of osmotic pressure of these solutions is:

Official previous-year question

Held on 30 Apr 2021 · Verified 9 Jul 2026.

Options

A
${P}_{2}>{P}_{1}>{P}_{3}$
B
${P}_{1}>{P}_{2}>{P}_{3}$
C
${P}_{2}>{P}_{3}>{P}_{1}$
D
${P}_{3}>{P}_{1}>{P}_{2}$

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Solution

Osmotic pressure $(\pi )=\mathrm{CRT}$

At same temperature, $\pi \propto C$

Solution: $(1)$ ${C}_{1}=\frac{10}{180}=\frac{10}{180}\times 4$

${C}_{1}=\frac{250}{1000}=\frac{40}{180}=\frac{2}{9}$

${C}_{1}=0.222M$

Solution: $(2)$ ${C}_{2}=\frac{10}{\frac{60}{\frac{250}{1000}}}=\frac{1}{6}\times 4=0.667M$

Solution: $(3)$ ${C}_{3}=\frac{10}{342}\times \frac{1000}{250}=0.1169M$

$\because {C}_{2}>{C}_{1}>{C}_{3}\therefore$ Hence, order: ${P}_{2}>{P}_{1}>{P}_{3}$

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Solution

Osmotic pressure $(\pi )=\mathrm{CRT}$

At same temperature, $\pi \propto C$

Solution: $(1)$ ${C}_{1}=\frac{10}{180}=\frac{10}{180}\times 4$

${C}_{1}=\frac{250}{1000}=\frac{40}{180}=\frac{2}{9}$

${C}_{1}=0.222M$

Solution: $(2)$ ${C}_{2}=\frac{10}{\frac{60}{\frac{250}{1000}}}=\frac{1}{6}\times 4=0.667M$

Solution: $(3)$ ${C}_{3}=\frac{10}{342}\times \frac{1000}{250}=0.1169M$

$\because {C}_{2}>{C}_{1}>{C}_{3}\therefore$ Hence, order: ${P}_{2}>{P}_{1}>{P}_{3}$

NEET UG 2021 — Chemistry Physical Chemistry

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NEET UG 2021 — Chemistry Physical Chemistry

Options

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Solution

Your note

NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry